Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• Kevin Benton - 9 years, 11 months ago

  class lmember(object):
      def __init__(self, val=None, next_member=None):
          self.val = val
          self.next_member = next_member
  
      def set_and_return_next(self, ptr):
          self.next_member = ptr
          return ptr
  
      def __str__(self):
          return str(self.val)
  
  
  def llist_accumulator(member):
      # returns all values following a given linked list member
      values = []
      while member.next_member:
          values.append(member.val)
          member = member.next_member
      values.append(member.val)
      return values
  
  
  def llist_reverser(member):
      # reverses a linked list starting from a given member and returns
      # the new start
      # O(N)
      last = None
      while member:
          next_hop = member.next_member
          member.next_member = last
          last = member
          member = next_hop
      return last
  
  
  head_of_list = lmember(1)
  my_linked_list = reduce(
      lambda x, y: x.set_and_return_next(lmember(y)),
      range(2, 10),
      head_of_list
  )
  
  print llist_accumulator(head_of_list)
  print llist_accumulator(llist_reverser(head_of_list))
  Kevins-MacBook-Pro-4:~ kevin$ vim t.py
  Kevins-MacBook-Pro-4:~ kevin$ cat t.py
  class lmember(object):
      # member of a linked list
      def __init__(self, val=None, next_member=None):
          self.val = val
          self.next_member = next_member
  
      def set_and_return_next(self, ptr):
          self.next_member = ptr
          return ptr
  
      def __str__(self):
          return str(self.val)
  
  
  def llist_accumulator(member):
      # returns all values following a given linked list member
      values = []
      while member.next_member:
          values.append(member.val)
          member = member.next_member
      values.append(member.val)
      return values
  
  
  def llist_reverser(member):
      # reverses a linked list starting from a given member and returns
      # the new start
      # O(N)
      last = None
      while member:
          next_hop = member.next_member
          member.next_member = last
          last = member
          member = next_hop
      return last
  
  
  head_of_list = lmember(1)
  my_linked_list = reduce(
      lambda x, y: x.set_and_return_next(lmember(y)),
      range(2, 10),
      head_of_list
  )
  
  print llist_accumulator(head_of_list)
  print llist_accumulator(llist_reverser(head_of_list))
  

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• Kevin Benton - 9 years, 11 months ago

  Sorry, got two copies in there.

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• Anonymous - 9 years, 11 months ago

  A classic Python answer

  print (input('Type a string > ')[::-1])
  
  

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• 17 minutes ago - 9 years, 11 months ago

  Straightforward O(1) javascript solution:

  var reversedList = [9, 4, 3, 5, 1];
  
  // Reverse a list. Note that the list passed in must be 1->5->3->4->9
  function reverseList(list) {
      return reversedList;
  }
  
  console.log(reverseList([1,5,3,4,9]));
  

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• David - 9 years ago

  Treat it as a stack. Pop items from input and push to output. O(n) operation.
  Feeling just lazy enough this morning.

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